First, note that the domain of #ln(x)# is #x>0#. Then, the domain of #ln(x-k)# is #x-k>0#, or #x>k#. We can neglect the #+2# in determining this because it has no bearing on domain--it constitutes a vertical shift.

Furthermore, note that #ln(x)# is asymptotic at #x=0#. Symbolically, #lim_(xrarr0^+)ln(x)=-oo#. Here, this becomes:

- #lim_(xrarrk^+)ln(x-k)=-oo#

One other valuable thing to note is that #ln(x)=0# at #x=1#. Then, #ln(x-k)=0# at #x-k=1#, or at #x=k+1#.

- #ln(x-k)# has a zero at #x=k+1#.

Assuming we know the general shape of the logarithmic curve (which for #ln(x)# is a reflection of the exponential curve #e^x# across the line #y=x#), we can draw the function #ln(x-k)#.

Arbitrarily setting #k=5# for the sake of graphing the function in the Socratic graphing system, we know that #ln(x-5)# will have the domain #x>5#, have an asymptote to #-oo# at #x=5# and have a zero at #x=5+1=6#. This looks like:

#ln(x-5)#

graph{ln(x-5) [1.01, 21.01, -6, 4]}

The missing piece here is that the function we want is #ln(x-k)+2#. So, all we do is we take each point on the graph of #ln(x-k)# and shift it up #2#. Again using our test function with #k=5#:

#ln(x-5)+2#

graph{ln(x-5)+2 [2.76, 22.76, -4.28, 5.72]}

If we're unsatisfied with this process of shifting vertically, we can also find the formal point at which #ln(x-k)+2# has a zero. This is when #ln(x-k)=-2#, so #x=k+e^-2#.

When #k=5#, this translates to a zero at #x~~5.1353#.